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A certain characterstic in a large population has a

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Manager
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A certain characterstic in a large population has a [#permalink] New post 08 Sep 2004, 03:39
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A certain characterstic in a large population has a distribution that is symmetric about the mean M. If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distribution is less than M+D?
A. 16% B.32% C. 48% D. 84% E. 92%

Please explain.
Thanks.
Manager
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 [#permalink] New post 08 Sep 2004, 09:14
That's just your straight normal distribution.

Symmetry=> 50% of the distribution lies above the mean (M) and 50% below

If 68% lies within 1 s.d. (D) of M, then 34% lies within +1 D and 34% lies within -1 D.

Thus, the distribution below M+D is the same as the complement of the distribution above M+D, or 1- 0.16 = 0.84.

Alternatively, the distribution below can be found as the sum of the distributions b/w M and +D (34%), M and -D (34%), and below -D (16%), which also sums to 84%.

Answer is D.
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 [#permalink] New post 08 Sep 2004, 09:59
Illustrated below for easier understanding..
Attachments

SD.JPG
SD.JPG [ 32.94 KiB | Viewed 409 times ]


Last edited by venksune on 08 Sep 2004, 12:52, edited 1 time in total.
Joined: 31 Dec 1969
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 [#permalink] New post 08 Sep 2004, 10:43
A certain characterstic in a large population has a distribution that is symmetric about the mean M. If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distribution is less than M+D?
A. 16% B.32% C. 48% D. 84% E. 92%

If we could draw a normal bell shaped distribution, we could make the explanation clearer.

In the absence of the normal distribution let me improvise it with this dotted line


A-------- (M-D)----------M------------(M +D)-----------B
In this diagram M is the mean D represents the length of a deviation
This implies that (M-D) to (M+D) represents distribution lying within one standard deviation of the mean and this is 68%
The distribution lying within A to (M-D) and that lying between (M+D) to B must equally account for the remaining 32% (100-68) of the disdtribituion
Thus each of these points will have 16% of the distribution

So we have
A to (M-D) = 16%
(M-D) to M +D =68%
(M+D) to B = 16%

We are intereted in distribution less than (M+D) ie the distribution from A to (M +D) = 16% + 68% = 84%

I hope this help
Manager
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 [#permalink] New post 08 Sep 2004, 23:23
Thanks everyone. I understand now.
Venksune,
This distribution - 68% , 95% and 99% will hold true only for a normal distribution is it. Is there any other concept similar to this in SD and distributions.
Thanks.
  [#permalink] 08 Sep 2004, 23:23
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